Surprising Perpendicularity - Problem 1

In $\triangle ABC$ with $\angle ABC=90^\circ$ , let $D$ and $E$ be the intersection of the angle bisectors of $\angle CAB$ and $\angle BCA$ with $BC$ and $AB$ , respectively. If $I$ is the intersection of $AD$ and $CE$ , and $O$ is the circumcenter of $\triangle BED$ , show that $OI\perp CA$ .

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Comments

Let $D',J,E'$ be the projections of $D,I,E$ on $AC$ . Since $O$ is just the midpoint of $ED$ , we only have to prove $JD'=JE'$ . By the bisector and Pythagoras' theorems we have: $CD'=CD\cos C=CD\sin A=\frac{ab}{b+c}\sin A=\frac{a^2}{b+c}=\frac{a^2(b-c)}{b^2-c^2}=b-c,$ and $AE'=b-a$ . Since $AJ=p-a$ and $CJ=p-c$ , the claim follows.

Jack D'Aurizio - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$